Question

the 2nd term of an AP is 9 and the 4th term is 25. Find the sum of the first 30 terms.​

Answer 1

$\color {red}\huge\bold\star\underline\mathcal{Question:-}$ we have to find sum of first 30 terms ..

$\huge\underline\blue{\sf Given:}$ a(2) = 9 & a(4) = 25

$\bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\:Answer}}}}}}}}}}$

we know That ,

a(n) = a + (n-1)d

so,

a(2) = a + d = 9 ----------------- Equation (1)

a(4) = a+3d = 25 --------------- Equation (2)

Subtracting Equation(1) From Equation(2) we get,

2d = 16

d = 8

putting in any now , we get , a = 1

so, now we have to find sum of 30 terms .

we also know that ,

$\large\red{\boxed{\sf Sn = \frac{n}{2}[2a + (n - 1)d]}}$

Putting values we get,

$s(30) = \frac{30}{2}(2 \times 1 + 29 \times 8) \\ \\ \\ s(30) = 15(2 + 232) \\ \\ \\ s(30) = 3510$

$\huge\blue{THANKS}$

Answer 2

Step-by-step explanation:

we know That ,

a(n) = a + (n-1)d

so,

a(2) = a + d = 9 ----------------- Equation (1)

a(4) = a+3d = 25 --------------- Equation (2)

Subtracting Equation(1) From Equation(2) we get,

2d = 16

d = 8

putting in any now , we get , a = 1

so, now we have to find sum of 30 terms .

we also know that ,

\large\red{\boxed{\sf Sn = \frac{n}{2}[2a + (n - 1)d]}}

Sn=

2

n

[2a+(n−1)d]

Putting values we get,

\begin{lgathered}s(30) = \frac{30}{2}(2 \times 1 + 29 \times 8) \\ \\ \\ s(30) = 15(2 + 232) \\ \\ \\ s(30) = 3510\end{lgathered}

s(30)=

2

30

(2×1+29×8)

s(30)=15(2+232)

s(30)=3510

\huge\blue{THANKS}